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Kepler's third law - semi-major axis

Posted on 2006-11-28
Medium Priority
Last Modified: 2012-06-21
In Kepler's third law (T² = 4pi²/GM r³), the r is the semi-major axis.

But why?

Why not the semi-minor axis? Or the average radius throughout the eccentric orbit?

(I'd be happy with a mathematical answer if that's easier)
Question by:InteractiveMind
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Assisted Solution

GnarOlak earned 100 total points
ID: 18030957
Math class was a long time ago so I don't even pretend to understand a lot of this but's_laws_of_planetary_motion has a proof of Kepler's Third Law.  Hope it helps.

Expert Comment

ID: 18030968
Clicking that link won't work.  EE truncated the link at the apostrophe.  Copy the entire URL into a browser.
LVL 27

Assisted Solution

d-glitch earned 100 total points
ID: 18031863
A planet is at the ends of the semi-major axis when it is closest to the sun (max kinetic energy)
and when it is furthest from the sun (max potential energy).

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LVL 84

Assisted Solution

ozo earned 200 total points
ID: 18032280
if you imagine orbits with an eccentricity close to 1, it may be more intuitive that semi-major axis makes more sense there than semi-minor axis
LVL 22

Assisted Solution

grg99 earned 100 total points
ID: 18033003
becuz mathematically the semi-minor axis can go to zero, and division by zero leads to big trouble.

But the other axis can only get bigger, which is mathematically tenable.

LVL 44

Expert Comment

ID: 18036297
for a mathematical proof of the Third law (heavy on the algebra - just a warning), see:'s_laws_of_planetary_motion#Proving_Kepler.27s_third_law

(becuase of the ' in the URL, you will nedd to COPY (Ctrl-C) and Paste (Ctrl-V) the URL to the Browser, you cannot simply clink on the link that shows in this answer)

LVL 44

Accepted Solution

Arthur_Wood earned 500 total points
ID: 18036439
In addition, Kepler used the Semi-Major axis because that is the measured parameter that characterizes the orbit.  Semi-minor axis cannot be used to characterize the orbit, since it is dependent on the Eccentricity of the orbit.

Kepler's 3rd law has no dependency, whatsoever, of the eccentricity of the orbit.  A very eccentiric orbit will have the same period as a very circular orbit, if both have the same Semi-major axis.  

LVL 44

Assisted Solution

Arthur_Wood earned 500 total points
ID: 18036557
Kepler did not state the 3rd law as :

T² = 4pi²/GM r³

but rather as a relationship that states the the square of the Period is PROPORTIONAL to the cube of then Semi-major axis.  This was taken entirely from observations of the actual motions of the planets.  Remember that Kepler was trying to match the positions of the planets, in the sky, as observed very accurately by Tycho Brahe (Kepler was NOT an observational astronomer - he used the observations provided by others, principally Tycho Brahe in Denmark, about 20 years before Kepler joined him as an assistant.

The mathematical formulation came much later, with the derivation from Newton's Law of Gravity.
LVL 25

Author Comment

ID: 18040819
thank you all

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